# Determining when a function is an element of L^2 (0,1)

## Homework Statement

Determine for what k f(x)=xk is an element of L2 (0,1) vector space
k ∈ ℝ

## The Attempt at a Solution

$$\int_{0}^{1} x^{2k} dx = \frac{1-0^{2k+1}}{1+2k} = \sum_{n=0}^{\infty}{(-2k)^{n}}$$ (for k > -½)

This sum should converge for $$\lim_{n \to +\infty} {\frac{|(-2k)^{n+1}|}{|(-2k)^{n}|}} < 1 = |-2k| < 1$$

Which gives me a radius of convergence for
$$- \frac{1}{2} < k < \frac{1}{2}$$

But just by examining it, the integral should exist for any k greater than negative one-half, what is wrong with my ratio test?

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LCKurtz
Homework Helper
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## Homework Statement

Determine for what k f(x)=xk is an element of L2 (0,1) vector space
k ∈ ℝ

## The Attempt at a Solution

$$\int_{0}^{1} x^{2k} dx = \frac{1-0^{2k+1}}{1+2k} = \sum_{n=0}^{\infty}{(-2k)^{n}}$$ (for v > -½)

This sum should converge for $$\lim_{n \to +\infty} {\frac{|(-2k)^{n+1}|}{|(-2k)^{n}|}} < 1 = |-2k| < 1$$

Which gives me a radius of convergence for
$$- \frac{1}{2} < k < \frac{1}{2}$$

But just by examining it, the integral should exist for any k greater than negative one-half, what is wrong with my ratio test?
The infinite series doesn't represent the fraction for all values of k. The series may not converge for some k but that doesn't mean the fraction's value doesn't exist.

Ok thanks, obvious mistake

HallsofIvy
I don't see your logic. The integral of $x^{2k}$ is the number 1/(2k+ 1) as long as that number exists. What does that have to do with whether or not there is a geometric sequence converging to it? If k= 1, which is greater than 1/2, f(x)= x which is certainly twice integrable between 0 and 1.